Question

Source

• leetcode: Find Minimum in Rotated Sorted Array II | LeetCode OJ
• lintcode: (160) Find Minimum in Rotated Sorted Array II

Problem

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

The array may contain duplicates.

Example

Given [4,4,5,6,7,0,1,2] return 0

题解

由于此题输入可能有重复元素，因此在 num[mid] == num[end] 时无法使用二分的方法缩小 start 或者 end 的取值范围。此时只能使用递增 start/递减 end 逐步缩小范围。

C++

class Solution {
public:
  /**
   * @param num: a rotated sorted array
   * @return: the minimum number in the array
   */
  int findMin(vector<int> &num) {
    if (num.empty()) {
      return -1;
    }

    vector<int>::size_type start = 0;
    vector<int>::size_type end = num.size() - 1;
    vector<int>::size_type mid;
    while (start + 1 < end) {
      mid = start + (end - start) / 2;
      if (num[mid] > num[end]) {
        start = mid;
      } else if (num[mid] < num[end]) {
        end = mid;
      } else {
        --end;
      }
    }

    if (num[start] < num[end]) {
      return num[start];
    } else {
      return num[end];
    }
  }
};

Java

public class Solution {
  /**
   * @param num: a rotated sorted array
   * @return: the minimum number in the array
   */
  public int findMin(int[] num) {
    if (num == null || num.length == 0) return Integer.MIN_VALUE;

    int lb = 0, ub = num.length - 1;
    // case1: num[0] < num[num.length - 1]
    // if (num[lb] < num[ub]) return num[lb];

    // case2: num[0] > num[num.length - 1] or num[0] < num[num.length - 1]
    while (lb + 1 < ub) {
      int mid = lb + (ub - lb) / 2;
      if (num[mid] < num[ub]) {
        ub = mid;
      } else if (num[mid] > num[ub]){
        lb = mid;
      } else {
        ub--;
      }
    }

    return Math.min(num[lb], num[ub]);
  }
}

源码分析

注意 num[mid] > num[ub] 时应递减 ub 或者递增 lb.

复杂度分析

最坏情况下 O(n)O(n)O(n), 平均情况下 O(logn)O(\log n)O(logn).